>>223
∫[x=0〜π]√{1-sin^2(x)} dx=∫[x=0〜π]|cos(x)| dx=∫[x=0〜π/2]cos(x) dx -∫[x=π/2〜π]cos(x)dx
={sin(π/2)-sin(0)}-{sin(π)-sin(π/2)}=1+1=2